Question

A system being studied in a 3.0-liter reactor was charged with 0.150 moles of H2, 0.150 moles of X2, and 0.600 moles of HX. A catalyst was introduced using a remote unit. Is the reaction at equilibrium

Answer 1

Answer:

The reaction is not in equilibrium and it proceed in reverse direction to reach the equilibruim condition

Explanation:

We have

H₂(g) + x₂(g)  ⇆  2HX (g)

Kc = 2.4 x 10 ⁻³ at 25°C

$K_c = \frac{[HX]^2}{[H_2][X_2]}$

Now we have 3L reactor and the number of mole of species are 0.150 mole H₂, 0.150 mole X₂ and 0.600 mole HX

Therefore,

$[H_2] = \frac{0.150}{3} \\= 0.05M$

$[X_2] = \frac{0.150}{3} \\= 0.05M$

$[HX] = \frac{0.600}{3} \\= 0.2M$

Therefore,

$Q_c = \frac{[HX]^2}{[H_2][X_2]}$

$Q_c = \frac{[0.200]^2}{[0.05][0.05]}\\\\=16$

Now, we calculate the free energy change for the reaction

ΔrG = RT In Q/K

$= (8.314) \times (298) \times In (\frac{16}{2.4\times 10^-^3} )\\\\=+21814.7 J/mol$

As free energy is +ve so the reaction is not in equilibruim and proceed in reverse reaction to reach tje equilibruim condition